Pisano period upper-bound for Tribonacci (3 step Fibonacci)
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For the Pisano Period of a 2-step Fibonacci at modulo $n$ a common and simple upper bound, according to this list of open problems, is $n^2-1$. Is there a similar upper bound for the Pisano Period of the Tribonacci (3-step Fibonacci) at modulo $n$?
number-theory fibonacci-numbers
asked Dec 4 '18 at 15:23
AstheroxAstherox
31
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add a comment |
$begingroup$
For the Pisano Period of a 2-step Fibonacci at modulo $n$ a common and simple upper bound, according to this list of open problems, is $n^2-1$. Is there a similar upper bound for the Pisano Period of the Tribonacci (3-step Fibonacci) at modulo $n$?
number-theory fibonacci-numbers
asked Dec 4 '18 at 15:23
AstheroxAstherox
31
$endgroup$
add a comment |
$begingroup$
For the Pisano Period of a 2-step Fibonacci at modulo $n$ a common and simple upper bound, according to this list of open problems, is $n^2-1$. Is there a similar upper bound for the Pisano Period of the Tribonacci (3-step Fibonacci) at modulo $n$?
number-theory fibonacci-numbers
asked Dec 4 '18 at 15:23
AstheroxAstherox
31
$endgroup$
For the Pisano Period of a 2-step Fibonacci at modulo $n$ a common and simple upper bound, according to this list of open problems, is $n^2-1$. Is there a similar upper bound for the Pisano Period of the Tribonacci (3-step Fibonacci) at modulo $n$?
number-theory fibonacci-numbers
number-theory fibonacci-numbers
asked Dec 4 '18 at 15:23
AstheroxAstherox
31
asked Dec 4 '18 at 15:23
AstheroxAstherox
31
asked Dec 4 '18 at 15:23
AstheroxAstherox
31
asked Dec 4 '18 at 15:23
AstheroxAstherox
31
asked Dec 4 '18 at 15:23
AstheroxAstherox
31
31
add a comment |
add a comment |
1 Answer
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Yes, $n^3-1$.
Consider any constant-coefficient linear recursion of order $m$. The values mod $n$ are determined by any $m$-tuple of consecutive values mod $n$. There are $n^m$ possible $m$-tuples mod $n$, but if $(0,ldots, 0)$ occurs the other terms would have to be all $0$. Thus if the sequence is nontrivial there are at most $n^m-1$ distinct possible $m$-tuples, and so the period will have to be at most $n^m-1$.
answered Dec 4 '18 at 15:33
Robert IsraelRobert Israel
321k23210462
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Do you know if this bound is sharp in general? I know it's sharp if $n = p$ is prime; then you can consider a recurrence with characteristic polynomial the minimal polynomial of a generator of the multiplicative group of $mathbb{F}_{p^m}$.
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– Qiaochu Yuan
Dec 4 '18 at 20:25
1
$begingroup$
No, it won't be sharp. Suppose $n = p q$ where $p$ and $q$ are coprime. Then the sequence mod $p$ has period at most $p^m-1$ and the sequence mod $q$ has period at most $q^m-1$, so the sequence mod $pq$ has period at most $(p^m-1)(q^m-1) < (pq)^m - 1$.
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– Robert Israel
Dec 4 '18 at 20:28
add a comment |
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1 Answer
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$begingroup$
Yes, $n^3-1$.
Consider any constant-coefficient linear recursion of order $m$. The values mod $n$ are determined by any $m$-tuple of consecutive values mod $n$. There are $n^m$ possible $m$-tuples mod $n$, but if $(0,ldots, 0)$ occurs the other terms would have to be all $0$. Thus if the sequence is nontrivial there are at most $n^m-1$ distinct possible $m$-tuples, and so the period will have to be at most $n^m-1$.
answered Dec 4 '18 at 15:33
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Do you know if this bound is sharp in general? I know it's sharp if $n = p$ is prime; then you can consider a recurrence with characteristic polynomial the minimal polynomial of a generator of the multiplicative group of $mathbb{F}_{p^m}$.
$endgroup$
– Qiaochu Yuan
Dec 4 '18 at 20:25
1
$begingroup$
No, it won't be sharp. Suppose $n = p q$ where $p$ and $q$ are coprime. Then the sequence mod $p$ has period at most $p^m-1$ and the sequence mod $q$ has period at most $q^m-1$, so the sequence mod $pq$ has period at most $(p^m-1)(q^m-1) < (pq)^m - 1$.
$endgroup$
– Robert Israel
Dec 4 '18 at 20:28
add a comment |
$begingroup$
Yes, $n^3-1$.
Consider any constant-coefficient linear recursion of order $m$. The values mod $n$ are determined by any $m$-tuple of consecutive values mod $n$. There are $n^m$ possible $m$-tuples mod $n$, but if $(0,ldots, 0)$ occurs the other terms would have to be all $0$. Thus if the sequence is nontrivial there are at most $n^m-1$ distinct possible $m$-tuples, and so the period will have to be at most $n^m-1$.
answered Dec 4 '18 at 15:33
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Do you know if this bound is sharp in general? I know it's sharp if $n = p$ is prime; then you can consider a recurrence with characteristic polynomial the minimal polynomial of a generator of the multiplicative group of $mathbb{F}_{p^m}$.
$endgroup$
– Qiaochu Yuan
Dec 4 '18 at 20:25
1
$begingroup$
No, it won't be sharp. Suppose $n = p q$ where $p$ and $q$ are coprime. Then the sequence mod $p$ has period at most $p^m-1$ and the sequence mod $q$ has period at most $q^m-1$, so the sequence mod $pq$ has period at most $(p^m-1)(q^m-1) < (pq)^m - 1$.
$endgroup$
– Robert Israel
Dec 4 '18 at 20:28
add a comment |
$begingroup$
Yes, $n^3-1$.
Consider any constant-coefficient linear recursion of order $m$. The values mod $n$ are determined by any $m$-tuple of consecutive values mod $n$. There are $n^m$ possible $m$-tuples mod $n$, but if $(0,ldots, 0)$ occurs the other terms would have to be all $0$. Thus if the sequence is nontrivial there are at most $n^m-1$ distinct possible $m$-tuples, and so the period will have to be at most $n^m-1$.
answered Dec 4 '18 at 15:33
Robert IsraelRobert Israel
321k23210462
$endgroup$
Yes, $n^3-1$.
Consider any constant-coefficient linear recursion of order $m$. The values mod $n$ are determined by any $m$-tuple of consecutive values mod $n$. There are $n^m$ possible $m$-tuples mod $n$, but if $(0,ldots, 0)$ occurs the other terms would have to be all $0$. Thus if the sequence is nontrivial there are at most $n^m-1$ distinct possible $m$-tuples, and so the period will have to be at most $n^m-1$.
answered Dec 4 '18 at 15:33
Robert IsraelRobert Israel
321k23210462
answered Dec 4 '18 at 15:33
Robert IsraelRobert Israel
321k23210462
answered Dec 4 '18 at 15:33
Robert IsraelRobert Israel
321k23210462
answered Dec 4 '18 at 15:33
Robert IsraelRobert Israel
321k23210462
321k23210462
$begingroup$
Do you know if this bound is sharp in general? I know it's sharp if $n = p$ is prime; then you can consider a recurrence with characteristic polynomial the minimal polynomial of a generator of the multiplicative group of $mathbb{F}_{p^m}$.
$endgroup$
– Qiaochu Yuan
Dec 4 '18 at 20:25
1
$begingroup$
No, it won't be sharp. Suppose $n = p q$ where $p$ and $q$ are coprime. Then the sequence mod $p$ has period at most $p^m-1$ and the sequence mod $q$ has period at most $q^m-1$, so the sequence mod $pq$ has period at most $(p^m-1)(q^m-1) < (pq)^m - 1$.
$endgroup$
– Robert Israel
Dec 4 '18 at 20:28
add a comment |
$begingroup$
Do you know if this bound is sharp in general? I know it's sharp if $n = p$ is prime; then you can consider a recurrence with characteristic polynomial the minimal polynomial of a generator of the multiplicative group of $mathbb{F}_{p^m}$.
$endgroup$
– Qiaochu Yuan
Dec 4 '18 at 20:25
1
$begingroup$
No, it won't be sharp. Suppose $n = p q$ where $p$ and $q$ are coprime. Then the sequence mod $p$ has period at most $p^m-1$ and the sequence mod $q$ has period at most $q^m-1$, so the sequence mod $pq$ has period at most $(p^m-1)(q^m-1) < (pq)^m - 1$.
$endgroup$
– Robert Israel
Dec 4 '18 at 20:28
$begingroup$
Do you know if this bound is sharp in general? I know it's sharp if $n = p$ is prime; then you can consider a recurrence with characteristic polynomial the minimal polynomial of a generator of the multiplicative group of $mathbb{F}_{p^m}$.
$endgroup$
– Qiaochu Yuan
Dec 4 '18 at 20:25
$begingroup$
Do you know if this bound is sharp in general? I know it's sharp if $n = p$ is prime; then you can consider a recurrence with characteristic polynomial the minimal polynomial of a generator of the multiplicative group of $mathbb{F}_{p^m}$.
$endgroup$
– Qiaochu Yuan
Dec 4 '18 at 20:25
1
1
$begingroup$
No, it won't be sharp. Suppose $n = p q$ where $p$ and $q$ are coprime. Then the sequence mod $p$ has period at most $p^m-1$ and the sequence mod $q$ has period at most $q^m-1$, so the sequence mod $pq$ has period at most $(p^m-1)(q^m-1) < (pq)^m - 1$.
$endgroup$
– Robert Israel
Dec 4 '18 at 20:28
$begingroup$
No, it won't be sharp. Suppose $n = p q$ where $p$ and $q$ are coprime. Then the sequence mod $p$ has period at most $p^m-1$ and the sequence mod $q$ has period at most $q^m-1$, so the sequence mod $pq$ has period at most $(p^m-1)(q^m-1) < (pq)^m - 1$.
$endgroup$
– Robert Israel
Dec 4 '18 at 20:28
add a comment |
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